CIS 307: Files and all that

These notes address material covered in Chapters 4 and 5 of Tannenbaum's textbook (and a little bit from chapters 7 and 11). It is assumed that you will study chapters 4 and 5, finish studying chapter 7. Of course, you have already studied file structures and basic IO operations in CIS207

Accessing a block

Seek Time

This is the time for moving the read/write head to the correct cylinder. The head will accelerate at the beginning, move at constant speed for a while, then decelerate. We can think of this time as a + b*d, where a and b are disk specific constants and d is the distance in cylinders from the start to the target cylinder. This seek time is on average many milliseconds, even many tens of milliseconds. Let's say it is 10ms. We may try to improve the seek time by carefully sheduling disk operations, or by locating records of a file so that the distance between them is not great.

Rotational Delay (Latency)

This is the time, once we are at the right cylinder, to wait for the sector to come to the read/write head. If we are lucky, it is immediately there. If unlucky we have to wait for a whole revolution. On average we will wait 1/2 a revolution. Since a disk spins fast, but not too fast [say, 6,000rpm, or 7,200rpm], a revolution will take about 10ms and the rotational delay will be about 5ms. We can do little to reduce the rotational delay [we may use disk scheduling, or use multiple heads, or improve sequential accesses, or ...].

Transfer Time

This is the time required to transfer the information. If we are transferring k bytes and the transfer rate is r bytes per second, this time is k/r. The transfer rate will be the smaller of two rates, the transfer rate of the bus to which the disk is connected [say, 10MBps and up], and the transfer rate of the disk [i.e. what corresponds to the rotation of the disk]. Say that a track has 64KB and the rotational speed is 6000rpm, then this data rate is 6.4MBps]. Usually the transfer rate of the disk is the smaller of the two trasfer rates. Note that as data density increases on disk, the disk data transfer rate increases in proportion If we are transferring an 8KB block then the transfer time is 1.125ms (in general, if we transfer k bytes, a track contains n bytes, and a rotation takes r milliseconds, then the transfer time is k*r/n). Unless we use very big blocks, the transfer time is much smaller than the the seek and rotational delays.

Stable Storage

We consider two disks D1 and D2 that are mirror images of each other, i.e. they have the same number of equally numbered sectors. Corresponding sectors are intended to have the same content. Here is how we write a value B to a sector s that has currently value A:

 1
    REPEAT
 2      write B to sector s on D1;
 3      read sector s from D1;
 4  UNTIL value read is without error;
 5
    REPEAT
 6      write B to sector s on D2;
 7      read sector s from D2;
 8  UNTIL value read is without error;
 9

We have numbered significant locations in the code. At 1 disks D1 and D2 have the same value A at sector s. At 2, 3, 4, and 5 D2 has value A at s. At 5 D1 has B at s and D2 has A at s. At 6, 7, and 8 D1 at s has value B. Reasoning similarly we see that at 9 we will have value B at s in both D1 and D2. Thus if all goes well we have succeeded in making D1 and D2 mirrors of each other. The problem is when there is a failure at an intermediate point after 1 and before 9. Here is the procedure to follow after a failure.

 10 read sector s from D1;
    if read error then {
      REPEAT
        read value of sector s from D2 and write to sector s on D1;
        read sector s from D1;
      UNTIL value read is without error;
      return;}
 11 read sector s from D2;
    if read error then {
      REPEAT
        read value of sector s from D1 and write to sector s on D2;
        read sector s from D2;
      UNTIL value read is without error;
      return;}
    Compare values read from D1 and D2;
    if different then 
      REPEAT 
        write value read from s on D1 to s on D2;
        read sector s from D2;
      UNTIL value read is without error;

Data and Index Blocks

Some observations:

• Access time is different at different locations in a file. The first 10 blocks are accessed with a single read (the pointers are in main memory where the inode is brought when the file is opened). The next 1K blocks require up to two reads, one for the index block, one for the data block. The next 1M blocks require up to three reads, and the next 1G blocks require up to four reads. Clearly access time is good at the beginning of the file and gets worse and worse as we move towards its end. This is true also during sequential access [not good if we keep movies on disk].
• One needs to have a clear idea of how addressing is done, i.e. how to determine, given an address, which blocks to read and how to compute displacements. For example, if we seek for byte 100,000,000 in a file, we need to go to the 12,207 block, and then look for the 256th byte. We have to use the 12th inode pointer, thus skipping (10+1K) blocks and remaining with 11,173 blocks. From the 12th inode pointer we go to a double indirection index block. Here each pointer counts as 1K blocks. Thus we have to skip 10 pointers (i.e. 10*1K=10,240 blocks, thus remaining with 933 blocks to go) and follow the 11th pointer to a single indirection index block. Here we go to the 933 pointer and follow it to a data block where the 256th byte is what we are aiming for. Clearly we can write a procedure to do this calculation and to allocate index and data blocks if it is necessary for a write operation.
• The following is a legal sequence of operations: write one byte, seek 1MB(from beginning of file), write another byte, seek 1GB (from beginning of file) and write another byte. Most of the resulting file is with undefined content. One needs to understand the resulting structure of the file in terms of its data and index blocks. In the example we will have:
  • The first inode pointer points to a block where the first byte is written.
  • The next 9 inode pointers are null.
  • The next inode pointer points to an index block with 1014 null pointers, followed by a pointer to a data block where only the first byte is defined.
  • The next inode pointer points to an index block where the first 1023 pointers are null, etc.
  • The next inode pointer is undefined.
• File creation is a time consuming operation. We have to create an inode in the inode list, we have to create an entry in the directory where the file is located and, if the file is non empty, say it has the first byte, we need to allocate and initialize a data block. In general when a file is modified we may have to update directory, inode, index blocks, and data blocks. If these data structure are far apart we may spend a lot of time in seeking operations.
• Directories in Unix are like regular files. The only difference is that they contain file entries. Each file entry has two parts, the name of the file, and the inode number of the file. The latter is used to index into the in disk inode list and to retrieve the inode that describes the file. This inode will be in the in memory inode map while the file is open.

Blocks, Fragments, Sectors

No overwrites

And the following figure describes a volume in which we have two files, one with two blocks and the other with three.

When a volume is mounted we will have in (main) memory structures corresponding to the root (the root block is also called the superblock), c-root, to the catalog, c-catalog, and to the index blocks of each open file. The file system will be seen in memory from these structures. The following figure describes the state of volume and memory when the volume is mounted and we have opened the file f2.

We can describe the basic operations of this file system:

Mount()

1. Allocate a block in (main) memory, call it c-root, and copy root block to it
2. Allocate a block in memory, call it c-catalog, and copy catalog block to it
3. Set c-root[master] to the address of c-catalog.
4. Determine list of free blocks on disk (you could trace from root)

create(f): f is a file that does not exist (after, f is not open)

1. Allocate and initialize a free block on disk
2. Set c-catalog[f] to its address

[This operation could be made more efficient, to not require I/O]

open(f): f is a file that is not open

1. Allocate a block in memory, call it c-f, and copy index block f to it
2. Set c-root[f] to the address of c-f

read(f,p): f is an open file, p is one of its pages

1. Allocate a block in memory, call it c-p, and copy block p to it

write(f,p,v): f is an open file, p is a page of f, v is a page-content

1. Allocate a free block p' on disk and write v to it
2. Set c-f[p] to the address of p'

close(f): f is an open file

1. Allocate a free block f' on disk and copy c-f to it
2. Set c-catalog[f] to f'
3. Set c-root[f] to undefined

delete(f): f is an existing closed file

1. Take f out of c-catalog

dismount()

1. Close each open file
2. Allocate a free block catalog' on disk
3. Set c-root[master] and c-catalog[master] to catalog'
4. Write c-catalog on catalog'
5. Overwrite root with c-root. [Remember that root is in stable storage]

This technique is also efficient. We have seen that delete involves no I/O. The copy operation is also very fast because we only need to copy the index block. The data blocks are changed only for write operations.
[You may compare this approach to the way Unix maintains file consistency (Tannenbaum pp175-177).]

Read vs Write Efficiency

1. the low data rate on the main memory channel, and
2. the slowness of disk write operations.

Log structured files

The key idea of the log-structured file system is to delay writes so that they can be made in big chunks, called segments, that include data blocks, index blocks, directory blocks, inodes, and inode list blocks (all inode list blocks are accessible from a single fixed location on disk). A segment is 1MB (or 0.5MB). A problem is how to manage space for segments and to deal with deletions in their content. The solution chosen is to compact the content of segments thus freeing pre-existing segments. This can be done by a background process.

Conclusion